Algebra Know-It-ALL

652 Worked-Out Solutions to Exercises: Chapters 11 to 19

and

y=−18/223

These are the same solutions we obtained in the chapter text.

4. For reference, here again is the first equation stated in Prob. 1:

− 4 x+ 2 y− 3 z= 5

The solutions for x and y can be plugged into this equation, and it can then be solved for

z in steps, as follows:

− 4 × (−209/223)+ 2 × (−18/223)− 3 z= 5

836/223− 36/223 − 3 z= 5

800/223− 3 z= 5

− 3 z= 5 − 800/223

− 3 z= 1,115 / 223 − 800/223

− 3 z= 315/223

z= (315/223) / (−3)

z=−105/223

This is the same solution we obtained in the chapter text.

5. The first and second revised equations we obtained in the section “Eliminate One
   Variable” were

− 4 x+ 2 y− 3 z= 5

and

2 x− 5 y−z=− 1

The two-variable equation in x and y that we derived from these, as a result of eliminating

the variable z, was

− 10 x+ 17 y= 8

If we take this equation together with the solution to Prob. 1, we have the two-by-two

linear system

− 37 x− 4 y= 35

and

− 10 x+ 17 y= 8